Golden Fibonacci

Let $F_n = a$ and $F_{n+1} = b$ . By repeating the formula $F_k = F_{k-1} + F_{k-2}$ for all integers $k \geq 3$ , we get $F_{n+2} = a+b$ , $F_{n+3} = a+2b$ , $\dots$ , $F_{n+8} = 13a + 21b$ .

From this, we can get $\begin{aligned} \lim_{n\to\infty} \frac{2F_{n+8}}{F_n} &= \lim_{n\to\infty} \frac{26a + 42b}{a} \\ &= \lim_{n\to\infty} 26 + 42 \frac{b}{a} \\ &= 26 + 42 \bigg( \lim_{n\to\infty} \frac{F_{n+1}}{F_n} \bigg) \end{aligned}$

It is well established that $\lim_{n\to\infty} \frac{F_{n+1}}{F_n} = \varphi = \frac{1+\sqrt{5}}{2}$ and thus we substitute this in and we get $\begin{aligned} \lim_{n\to\infty} \frac{2F_{n+8}}{F_n} &= 26 + 42 (\frac{1+\sqrt{5}}{2}) \\ &= 47 + 21\sqrt{5} \\ &= 55 -8 + 21\sqrt{5} \\ &= F_{10} - F_6 + F_8\sqrt{F_5} \end{aligned}$

This gives us a solution of $5 + 6 + 8 + 10 = \boxed{29}$

I wanted to point out that previously, I thought that the question asked for the answer to be of the from $F_{\alpha} + F_{\beta} + F_{\gamma}\sqrt{F_{\delta}}$ (without the minus sign). In this case, you can still get the right answer of 29 since $F_n + F_{n+2} = F_{n+3} - F_{n-1}$ for all integers $n > 1$ . I got $47 + 21\sqrt{5} = F_{10} - F_6 + F_8\sqrt{F_5} = F_7 + F_9 + F_8\sqrt{F_5}$ giving the final sum to be $5 + 7 + 8 + 9 = \boxed{29}$

Fibonacci sequence is linear recurrent sequence tha follows the Binet's formula: $F_n = \dfrac{\varphi^n - (-\varphi)^{-n}}{\sqrt{5}}$ , where $\varphi = \dfrac{1+\sqrt{5}}{2}$ , the golden ratio. Therefore, we have:

$\begin{aligned} \lim_{n \to \infty} \frac{2F_{n+8}}{F_n} & = \lim_{n \to \infty} \frac{2(\varphi^{n+8} - (-\varphi)^{-n-8})}{\varphi^n - (-\varphi)^{-n}} \\ & = \lim_{n \to \infty} \frac{2(\varphi^{n+8-n} - (-\varphi)^{-n-8-n})}{\varphi^{n-n} - (-\varphi)^{-n-n}} \\ & = \lim_{n \to \infty} \frac{2(\varphi^{8} - (-\varphi)^{-2n-8})}{1 - (-\varphi)^{-2n}} \\ & = 2\varphi^8 \end{aligned}$

Since $\varphi$ is a root of $x^2 - x - 1 = 0$ or $x^2=x+1$ , therefore,

$\begin{aligned} \varphi^2 & = \varphi + 1 \\ \varphi^4 & = (\varphi + 1)^2 = \varphi^2 + 2\varphi + 1 = 3\varphi + 2 \\ \varphi^8 & = (3\varphi + 2)^2 = 9\varphi^2 + 12\varphi + 4 = 21\varphi + 13 \end{aligned}$

$\begin{aligned} \Rightarrow \lim_{n \to \infty} \frac{2F_{n+8}}{F_n} & = 2\varphi^8 \\ & = 2 \left(21\varphi + 13 \right) \\ & = 2 \left(\frac{21(1+\sqrt{5}}{2} + 13 \right) \\ & = 21\sqrt{5} + 47 \\ & = 21\sqrt{5} + 55 - 8 \\ & = F_8 \sqrt{F_5} + F_{10} - F_6 \end{aligned}$

$\Rightarrow \alpha + \beta + \gamma + \delta = 8 + 5 + 10 + 6 = \boxed{29}$